Equation of tangent line to given curve...

[Becky4paws]

Find the equation of the tangent line to the given curve at specific point.

(1/x) - (1/y) = 2
(1/4, 1/2)

(x^-1) - (y^-1) = 2
(-x^-2) + (y^-2) = 0
y^-2 = x^-2
(1/y^2) = (1/x^2)
dy/dx = y^2/x^2

slope = (1/2)^2/(1/4)^2 = (1/4)/(1/16) = 4

y-1/2 = 4(x-1/4)
y=4x-1/2

I think I might actually have gotten this right....
You won't hear me say that very often...
But I guess I shouldn't count my points til they're plotted. :roll:

 

[skeeter]

your "technique" for implicit differentiation leaves me thinking you "lucked" into the correct derivative. Should look like this ...

\(\displaystyle \L \frac{d}{dx}[\frac{1}{x} - \frac{1}{y} = 2]\)

\(\displaystyle \L -\frac{1}{x^2} + \frac{1}{y^2}\frac{dy}{dx} = 0\)

\(\displaystyle \L \frac{1}{y^2}\frac{dy}{dx} = \frac{1}{x^2}\)

\(\displaystyle \L \frac{dy}{dx} = \frac{y^2}{x^2}\)

tangent line equation is o.k.